Question 375244
Simplify first,
{{{x^2-9y^2=(x-3y)(x+3y)}}}
{{{3x^2-10xy+3y^2=(x-3y)(3x-y)}}}
{{{x^3+27y^3=(x+3y)(x^2-3xy+9y^2)}}}
{{{x^2-6xy+9y^2 =(x-3y)^2}}}
So then,
{{{(x^2-9y^2)/(3x^2-10xy+3y^2)=((x-3y)(x+3y))/((x-3y)(3x-y))}}}
{{{(x^2-9y^2)/(3x^2-10xy+3y^2)=(x+3y)/(3x-y)}}}
and
{{{(x^3+27y^3)/(x^2-6xy+9y^2)= ((x+3y)(x^2-3xy+9y^2))/(x-3y)^2}}}
and finally,
{{{((x^2-9y^2)/(3x^2-10xy+3y^2))/((x^3+27y^3)/(x^2-6xy+9y^2))=((x+3y)/(3x-y))*((x-3y)^2/((x+3y)(x^2-3xy+9y^2)))}}}
.
.
{{{((x^2-9y^2)/(3x^2-10xy+3y^2))/((x^3+27y^3)/(x^2-6xy+9y^2))=highlight((x-3y)^2/((3x-y)(x^2-3xy+9y^2)))}}}